Let $m,n$ be two positive integer, $n>m$. I have trouble proving that $$ \sum_{k=m}^n \frac{\binom{1/2}{k-m}}{k \binom{-1/2}{k}}=\frac{\binom{-1/2}{n-m}}{m \binom{-1/2}{n}} $$ Any suggestions, please? Thank you very much.
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$\begingroup$ Use $\binom{x}{k}=\frac{x(x-1)\dots(x-k+1)}{k!}$ and simplify. $\endgroup$– Lutz LehmannJan 3, 2014 at 19:35
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$\begingroup$ Looks amusing. What's the source? $\endgroup$– Grigory MJan 3, 2014 at 20:12
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$\begingroup$ @GrigoryM, my teacher. $\endgroup$– MarkJan 3, 2014 at 23:26
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1 Answer
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You have a summand that does not depend on $n$ and you have a conjecture for general $n$ (which is clearly true for $m=n$), so it just remains to check the induction step:
$$\frac{\binom{-1/2}{n-m}}{m \binom{-1/2}{n}}+\frac{\binom{1/2}{n-m+1}}{(n+1) \binom{-1/2}{n+1}}=\frac{\binom{-1/2}{n-m+1}}{m \binom{-1/2}{n+1}},$$
which reduces to a polynomial identity after you cancel all common factors.